Answer
$$\frac{1}{6}\left( {3\pi - 8} \right)$$
Work Step by Step
$$\eqalign{
& \left( {\bf{a}} \right){\text{ Using a CAS }}\left( {{\text{Wolfram Alpha}}} \right){\text{ we obtain}} \cr
& \int_{ - \pi /4}^{\pi /4} {{{\tan }^4}x} dx = \frac{1}{6}\left( {3\pi - 8} \right) \cr
& \cr
& \left( {\bf{b}} \right)\int_{ - \pi /4}^{\pi /4} {{{\tan }^4}x} dx \cr
& {\text{Rewrite the integrand}} \cr
& \int_{ - \pi /4}^{\pi /4} {{{\tan }^4}x} dx = \int_{ - \pi /4}^{\pi /4} {{{\tan }^2}x{{\tan }^2}x} dx \cr
& {\text{Use the identity }}{\tan ^2}x = {\sec ^2}x - 1 \cr
& \int_{ - \pi /4}^{\pi /4} {{{\tan }^2}x{{\tan }^2}x} dx = \int_{ - \pi /4}^{\pi /4} {{{\tan }^2}x\left( {{{\sec }^2}x - 1} \right)} dx \cr
& {\text{ }} = \int_{ - \pi /4}^{\pi /4} {{{\tan }^2}x{{\sec }^2}x} dx - \int_{ - \pi /4}^{\pi /4} {{{\tan }^2}x} dx \cr
& {\text{ }} = \int_{ - \pi /4}^{\pi /4} {{{\tan }^2}x{{\sec }^2}x} dx - \int_{ - \pi /4}^{\pi /4} {\left( {{{\sec }^2}x - 1} \right)} dx \cr
& {\text{ }} = \int_{ - \pi /4}^{\pi /4} {{{\tan }^2}x{{\sec }^2}x} dx - \int_{ - \pi /4}^{\pi /4} {{{\sec }^2}x} dx + \int_{ - \pi /4}^{\pi /4} {dx} \cr
& {\text{Integrating}} \cr
& = \left[ {\frac{{{{\tan }^3}x}}{3}} \right]_{ - \pi /4}^{\pi /4} - \left[ {\tan x} \right]_{ - \pi /4}^{\pi /4} + \left[ x \right]_{ - \pi /4}^{\pi /4} \cr
& = \left[ {\frac{{{{\tan }^3}x}}{3} - \tan x + x} \right]_{ - \pi /4}^{\pi /4} \cr
& {\text{Evaluating}} \cr
& = \left[ {\frac{{{{\tan }^3}\left( {\pi /4} \right)}}{3} - \tan \left( {\frac{\pi }{4}} \right) + \frac{\pi }{4}} \right] - \left[ {\frac{{{{\tan }^3}\left( { - \pi /4} \right)}}{3} - \tan \left( { - \frac{\pi }{4}} \right) - \frac{\pi }{4}} \right] \cr
& = \left[ {\frac{1}{3} - 1 + \frac{\pi }{4}} \right] - \left[ { - \frac{1}{3} + 1 - \frac{\pi }{4}} \right] \cr
& = - \frac{2}{3} + \frac{\pi }{4} - \frac{2}{3} + \frac{\pi }{4} \cr
& = \frac{\pi }{2} - \frac{4}{3} \cr
& = \frac{{3\pi - 8}}{6} \cr
& = \frac{1}{6}\left( {3\pi - 8} \right) \cr} $$
